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I have been confused about this question for a while now. If anyone could help, it would be really very helpful. The cumulative distribution function of the random variable $X$ is: $$ F(x)= \begin{cases} 0& \text{for }x<-1,\\ (x+1)/2& \text{for } -1\leq x<1,\\ 1& \text{for }x\geq1. \end{cases}$$

Find the probability density function of $X$. I am not sure but my guess is the derivative of cumulative distribution function gives you the probability density function???

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  • $\begingroup$ Yes, that's right. $\endgroup$ – Kenny Wong Apr 26 '17 at 15:48
  • $\begingroup$ Isn't this a theorem in your notes? $\endgroup$ – Did Apr 26 '17 at 18:10
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Every random variable $X$ has a distribution function $F(x)=\mathbb P(X\leq x)$, by definition. Also, by definition, if we're dealing with continuous random variables, we have: $$ F(x)=\mathbb P(X\leq x)=\int_{-\infty}^xf(t)dt, $$ where $f$ (non-negative, continuous) is our density function. Therefore, $$ f(x)=\frac{d}{dx}F(x), $$ where we're applying the Fundamental Theorem of Calculus.

Since your distribution function is defined in cases, you have to calculate the derivative in cases too. I will show one case: $F(x)=0$ for $x<-1$. Thus we have: $\frac{d}{dx}F(x)=0$ for $x<-1$. You should finish the other two cases, and then write out $f(x)$ in the same way as $F(x)$ has been written out, i.e. by using cases.

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  • $\begingroup$ Does this imply that for $x \ge 1 $, $f(x) = 0$ too? $\endgroup$ – Wallace Jan 11 at 20:07

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